Question 175 of 295

As shown in the figure, a combination of a thin plano-concave lens and a thin plano-convex lens is used to image an object placed at infinity. The radius of curvature of both the lenses is $30cm$ and refraction index of the material for both the lenses is $1.75$. Both the lenses are placed at distance of $40cm$ from each other. Due to the combination, the image of the object is formed at distance $x=$ ______ $cm$, from concave lens.

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Answer: 120

Using Lens Makers formula for the plano-concave part, we can write
$\dfrac{1}{{f}_{1}}=(1.75-1)\left(-\dfrac{1}{30}\right)$
$\Rightarrow {f}_{1}=-40cm$
Using Lens Makers formula for the plano-convex part, we can write
$\dfrac{1}{{f}_{2}}=\left(1.75-1\right)\left(\dfrac{1}{30}\right) \Rightarrow {f}_{2}=40cm$
Image from first lens will be virtual and on the left of first lens at focal length $40cm$. So the object for second lens will be $80cm$ from second lens which is $2{f}_{2}$. Final image is formed at $80cm$ from second lens on the right.
So, $x=120$.